

A032198


"CIK" (necklace, indistinct, unlabeled) transform of 1,2,3,4,...


3



1, 3, 6, 13, 25, 58, 121, 283, 646, 1527, 3601, 8678, 20881, 50823, 124054, 304573, 750121, 1855098, 4600201, 11442085, 28527446, 71292603, 178526881, 447919418, 1125750145, 2833906683, 7144450566, 18036423973
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OFFSET

1,2


LINKS

Table of n, a(n) for n=1..28.
C. G. Bower, Transforms (2)
P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 5860.
P. Flajolet and M. Soria, The Cycle Construction. [pdf file]
Index entries for sequences related to necklaces


FORMULA

a(n) = 1/n*Sum_{d divides n} phi(n/d)*A004146(d).  Vladeta Jovovic, Feb 15 2003
From Petros Hadjicostas, Jan 07 2018: (Start)
a(n) = 2 + (1/n)*Sum_{dn} phi(n/d)*A005248(d) = 2 + (1/n)*Sum_{dn} phi(n/d)*L(2*d), where L(n) = A000032(n) is the usual Lucas sequence.
G.f.: Sum_{n>=1} (phi(n)/n)*log(1B(x^n)), where B(x) = x + 2*x^2 + 3*x^3 + 4*x^4 + ... = x/(1x)^2.
G.f.: 2*x/(1x)  Sum_{n>=1} (phi(n)/n)*log(13*x^n+x^(2*n)).
(End)


EXAMPLE

From Petros Hadjicostas, Jan 07 2018: (Start)
We give some examples to illustrate the theory of C. G. Bower about transforms given in the weblink above. We assume we have boxes of different sizes and colors that we place on a circle to form a necklace. Two boxes of the same size and same color are considered identical (indistinct and unlabeled). We do, however, change the roles of the sequences (a(n): n>=1) and (b(n): n>=1) that appear in the weblink above. We assume (a(n): n>=1) = CIK((b(n): n>=1)).
Since b(1) = 1, b(2) = 2, b(3) = 3, etc., a box that can hold 1 ball only can be of 1 color only, a box that can hold 2 balls only can be one of 2 colors only, a box that can hold 3 balls can be one of 3 colors, and so on.
To prove that a(3) = 6, we consider three cases. In the first case, we have a single box that can hold 3 balls, and thus we have 3 possibilities for the 3 colors the box can be. In the second case, we have a box that can hold 2 balls and a box that can hold 1 ball. Here, we have 2 x 1 = 2 possibilities. In the third case, we have 3 identical boxes, each of which can hold 1 ball. This gives rise to 1 possibility. Hence, a(3) = 3 + 2 + 1 = 6.
To prove that a(4) = 13, we consider 5 cases: a box with 4 balls (4 possibilities), one box with 3 balls and one box with 1 ball (3 possibilities), two identical boxes each with 2 balls (3 possibilities), one box with 2 balls and two identical boxes each with 1 ball (2 possibilities), and four identical boxes each with 1 ball (1 possibility). Thus, a(4) = 4 + 3 + 3 + 2 + 1 = 13.
(End)


MATHEMATICA

nmax = 30;
f[x_] = Sum[n*x^n, {n, 1, nmax}];
gf = Sum[(EulerPhi[n]/n)*Log[1/(1  f[x^n])] + O[x]^nmax, {n, 1, nmax}]/x;
CoefficientList[gf, x] (* JeanFrançois Alcover, Jul 29 2018, after Joerg Arndt *)


PROG

(PARI)
N = 66; x = 'x + O('x^N);
f(x)=sum(n=1, N, n*x^n );
gf = sum(n=1, N, eulerphi(n)/n*log(1/(1f(x^n))) );
v = Vec(gf)
/* Joerg Arndt, Jan 21 2013 */


CROSSREFS

Cf. A000032, A004146, A005248, A032170.
Equals A005594(n)1.
Sequence in context: A215984 A074890 A244704 * A079941 A255125 A267367
Adjacent sequences: A032195 A032196 A032197 * A032199 A032200 A032201


KEYWORD

nonn


AUTHOR

Christian G. Bower


STATUS

approved



